The sum of 4 consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of the middle terms is 7:15. find the numbers.
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Let the four consecutive term of the AP be ,
a - 3 d , a - d , a + d , a + 3 d
A.T.Q.
a - 3 d + a - d + a + d + a + 3 d = 32
4 a = 32
a = 8
and
( a - 3 d ) ( a + 3 d ) / ( a - d ) ( a + d ) = 7 / 15
15 ( a² - 9 d² ) = 7 ( a² - d² )
8 a² = 128 d² [ a = 8 ]
d = ± 2
Therefore , numbers are 2 , 6 , 10 , 14 or 14 , 10 , 6 , 2 .
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